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$\sin \frac{23 π}{24} = \sqrt{\frac{2 \sqrt{p} - \sqrt{q} - 1}{4 \sqrt{r}}}$ then the value of $p^{2} - q^{2} - r^{2} =$

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Correct option is D

sin23π24=sinπ-π24 =sinπ24We have 2sin2π24=1-cosπ12 =1-cos15°∴sin2π24=1-3+1222⇒sinπ24=22-3-142⇒p=2,q=3 and r=2⇒p2+q2-r2=9

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sin23π24=2p−q−14rthen the value of p2−q2−r2=

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$\sin \frac{23 π}{24} = \sqrt{\frac{2 \sqrt{p} - \sqrt{q} - 1}{4 \sqrt{r}}}$ then the value of $p^{2} - q^{2} - r^{2} =$